Answer:
F=9/5(°C)+32
Explanation:
Lets take an example
- Take 25°C
An isolated system consists of a 1.5 kg mass moving in the presence of the following potential energy function: U open parenthes
es x close parentheses equals open parentheses 1 third straight J over straight m cubed close parentheses x cubed plus open parentheses 1 half straight J over straight m squared close parentheses x squared minus open parentheses 2 straight J over straight m close parentheses x What is the period (in s) of small amplitude oscillations of this system about its stable equilibrium point?
1 answer:
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Answer:
40.0⁰
Explanation:
The formula for calculating the magnetic flux is expressed as:
where:
is the magnetic flux
B is the magnetic field
A is the cross sectional area
is the angle that the normal to the plane of the loop make with the direction of the magnetic field.
Given
A = 0.250m²
B = 0.020T
= 3.83 × 10⁻³T· m²
3.83 × 10⁻³ = 0.020*0.250cosθ
3.83 × 10⁻³ = 0.005cosθ
cosθ = 0.00383/0.005
cosθ = 0.766
θ = cos⁻¹0.766
θ = 40.0⁰
<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>
Answer:
Explanation:
Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:
Where are linear kinetic energy, gravitational potential energy and work, respectively.